198 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

If you replace x with xi and sum over all i, you would get ∑mi=1 g(x1, · · · ,xm)xm−k

i whichwould also be a symmetric polynomial. It is of the form

g(x1, · · · ,xm)xm−k1 +g(x1, · · · ,xm)xm−k

2 + · · ·+g(x1, · · · ,xm)xm−km

so when you switch some variable in this, you get the same thing.Here is a very interesting result which I saw claimed in a paper by Steinberg and Red-

heffer on Lindermannn’s theorem which follows from the above theorem. It is a verysurprising property of symmetric polynomials (surprising for me anyway) and is the maintool for proving the Lindermann Weierstrass theorem.

Theorem 9.1.6 Let α1, · · · ,αn be roots of the polynomial equation

p(x)≡ anxn +an−1xn−1 + · · ·+a1x+a0 = 0 (∗)

where each ai is an integer. Then any symmetric polynomial in the quantities

anα1, · · · ,anαn

having integer coefficients is also an integer. Also any symmetric polynomial with rationalcoefficients in the quantities α1, · · · ,αn is a rational number.

Proof: Let f (x1, · · · ,xn) be the symmetric polynomial having integer coefficients.From Theorem 9.1.4 it follows there are integers ak1···kn such that

f (x1, · · · ,xn) = ∑k1+···+kn≤m

ak1···kn pk11 · · · p

knn (9.3)

where the pi are elementary symmetric polynomials defined as the coefficients of p̂(x) =∏

nj=1 (x− x j) with pk (x1, ...,xn) of degree k since it is the coefficient of xn−k. Earlier we

had them ± these coefficients. Thus

f (anα1, · · · ,anαn)

= ∑k1+···+kn=d

ak1···kn pk11 (anα1, · · · ,anαn) · · · pkn

n (anα1, · · · ,anαn)

Now the given polynomial in ∗, p(x) is of the form

an

n

∏j=1

(x−α j)≡ an

(n

∑k=0

pk (α1, · · · ,αn)xn−k

)

= anxn +an−1xn−1 + · · ·+a1x+a0

Thus, equating coefficients, an pk (α1, · · · ,αn) = an−k. Multiply both sides by ak−1n . Thus

pk (anα1, · · · ,anαn) = ak−1n an−k an integer. Therefore,

f (anα1, · · · ,anαn)

= ∑k1+···+kn=d

ak1···kn pk11 (anα1, · · · ,anαn) · · · pkn

n (anα1, · · · ,anαn)